Inside the proof of irrationality of $e^k$ for $k \in \mathbb{N}$ the book assumes the following without any details/proof:
If $f(z)=z^{p-1}(z-k)^p = \sum_{p-1}^{2p-1} c_N z^N$ then $\sum_{p-1}^{2p-1} |c_N|=(1+k)^p$.
I got $\sum_{p-1}^{2p-1} |c_N| \ge (1+k)^p$ not the equality, by $z=-1$. How $\sum_{p-1}^{2p-1} |c_N|=(1+k)^p$ holds, then?
By the binomial theorem, $$z^{p-1}(z-k)^p=\sum_{j=1}^p \binom{p}{j}(-1)^j k^j z^{2p-j-1}.$$ Therefore, also by the binomial theorem, $$\sum |c_N|=\sum \binom{p}{j}|k|^j=(1+|k|)^p.$$
Since $k\in\mathbb N$, $|k|=k$, and the desired equality follows.